Integrate the function: \(\frac {1}{(x^2+1)(x^2+4)}\)
Solution and Explanation
∴\(\frac {1}{(x^2+1)(x^2+4)}\) = \(\frac {Ax+B}{(x^2+1)}+\frac {Cx+D}{(x^2+4)}\)
\(⇒1 = (Ax+B)(x^2+4)+(Cx+D)(x^2+1)\)
\(⇒1 = Ax^3+4Ax+Bx^2+4B+Cx^3+Cx+Dx^2+D\)
Equating the coefficients of \(x^3,x^2,x,\) and constant term,we obtain
\(A+C=0\)
\(B+D=0\)
\(4A+C=0\)
\(4B+D=1\)
On solving these equations, we obtain
\(A=0,\ B=\frac 13,\ C=0,\ D=-\frac 13\)
From equation(1), we obtain
\(\frac {1}{(x^2+1)(x^2+4)}\) = \(\frac {1}{3(x^2+1)}-\frac {1}{3(x^2+4)}\)
\(∫\)\(\frac {1}{(x^2+1)(x^2+4)}\) = \(\frac 13∫\frac {1}{x^2+1}dx-\frac {1}3∫\frac {1}{x^2+4}dx\)
=\(\frac 13\tan^{-1}x-\frac 13.\frac 12tan^{-1}\frac x2+C\)
=\(\frac 13tan^{-1}x-\frac 16tan^{-1}\frac x2+C\)
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Concepts Used:
Integration by Partial Fractions
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.
For examples,
